Question

consider the following.
f(x) = (x)/(x^2-2)
Find the x-values at which f is not continuous,
smaller value =
larger value =
Is the discontinuity at the smaller value removable?
Is the discontinuity at the larger value removable?

Answer 1

look for a value of x that makes the denominator zero, it won't be continuous there

Answer 2

hint:
for continuity, we can not divide by zero

Answer 3

i'm confused on what to do...

Answer 4

we have tosolve this equation, first:
\[\Large {x^2} - 2 = 0\]

Answer 5

x=sqrt(2)

Answer 6

more precisely, we have two solutions:
\[\Large x = \pm \sqrt 2 \]

Answer 7

am I right?

Answer 8

yea. so the smaller value is \[-\sqrt{2}\] and the larger value is \[\sqrt{2} \]?

Answer 9

Correct!
we have to exclude both values, since we can not divide by zero, so our function is not continuous at these points:

\[\Large \begin{gathered}
x = \sqrt 2 \hfill \\
x = - \sqrt 2 \hfill \\
\end{gathered} \]

Answer 10

right. so how do you find out if it's removable?

Answer 11

I think that both discontinuities are not removable

Answer 12

can you explain why?

Answer 13

yes! I explain with an example

Answer 14

for example, the subsequent function:
\[\Large f\left( x \right) = \frac{{{x^2} - 1}}{{x - 1}}\]
is not continuous at x=1, right?

Answer 15

yes

Answer 16

nevertheless x=1 is a removable discontinuity, since we can write this:
\[\Large f\left( x \right) = \frac{{{x^2} - 1}}{{x - 1}} = \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} = x + 1\]

Answer 17

ooh ok. can you help me with another problem?

Answer 18

yes!

Answer 19

consider the following.
f(x) = {-9x

Answer 20

ok!

Answer 21

f(x) = -9 if x less than or equal to 5
x^2-2x+2 if x greater than 5

Answer 22

find the x-value at which f is not continuous and is the discontinuity removable

Answer 23

at x= 5, we have f(x)= 9 using your first part for f(x)
and we have:
\[{x^2} - 2x + 2 = 25 - 2 \cdot 5 + 2 = 17 \ne 9\]
so the discontinuity is at x=5

Answer 24

is it not removable?

Answer 25

I think that it is not removable

Answer 26

because it is a jump-type discontinuity

Answer 27

ok thanks!. I have a few more I need help me, is that ok?

Answer 28

ok!

Answer 29

g(x) =
5sin(x)/x if x less than 0
a-4x if x greater than or less than 0
Find the constant a such that the function is continuous on the entire real line.

Answer 30

we can write this:
\[\Large \mathop {\lim }\limits_{x \to 0} \frac{{5\sin x}}{x} = 5\]

Answer 31

whereas at x=0, we have:
\[\Large a - 4x = a - 4 \cdot 0 = a\]

Answer 32

now, for continuity, we have to request taht both values are equal each to other.
So what do you get?

Answer 33

Would the constant a be 5?

Answer 34

that*

Answer 35

correct! a=5

Answer 36

g(x) =
(x^2-a^2)/(x-a) if x is not equal to a
11 if x=a
Find the constant a such that the function is continuous on the entire real line.

Answer 37

we can write this:
\[\Large \frac{{{x^2} - {a^2}}}{{x - a}} = \frac{{\left( {x - a} \right)\left( {x + a} \right)}}{{x - a}} = x + a\]

Answer 38

and its value at x=a, is:
\[\Large 2a\]
Now, for continuity we have to request, as before, that those two values are equal each to other, so we have to solve this equation:
\[\Large 2a = 11\]

Answer 39

I don't understand how you got 2a

Answer 40

after a simplification, I got this:
\[\Large g\left( x \right) = x + a\]
so,
\[\Large g\left( a \right) = a + a = 2a\]

Answer 41

oh ok. so the constant a is 11/2?

Answer 42

yes! correct!

Answer 43

Describe the interval(s) on which the function below is continuous.
\[f(x)=x \sqrt{x+1}\]

Answer 44

your function is continuous inside all its domain, and its domain is the subset of all real numbers, such that the radicand is positive or equal to zero, namely:
\[\Large x + 1 \geqslant 0\]

Answer 45

[0,∞)
is that how you would write it?

Answer 46

we have:
\[\Large x + 1 \geqslant 0 \Rightarrow x \geqslant - 1\]
so, the requested interval is:
\[\Large [ - 1,\infty )\]
am I right?

Answer 47

yes. Thank you so much for the help and sorry if I took up your time.

Answer 48

no worries! :) :)